a particle A o mass m, moving with velocity u in the direction toward particle B,makes a head on elastic collision with a ststionary particle B of the mass M. after the collision,A and B move off with velocities v and V respectively.
a)write down the equations that summarise the application of the Principles of conservation Of Energy and Momentum to this collision.
b) it can be shown from these equation that v=V-u.Using this result ,or otherwise ,find an exprssion for the fractional loss of kinetic energy of A,in the terms of M and m only.
al phy
2009-11-15 10:13 pm
回答 (3)
2009-11-15 10:53 pm
✔ 最佳答案
(a) By conservation of linear momentum:mu = mv + MV ... (1)
By conservation of energy:
mu2/2 = mv2/2 + MV2/2
mu2 = mv2 + MV2 ... (2)
(b) With v = V - u:
Initial k.e. of A = mu2/2
Final k.e. of A = mv2/2
Hence fractional loss of k.e. of A is:
(mu2/2 - mv2/2)/(mu2/2)
= (u2 - v2)/u2
= (u - v)(u + v)/u2
= (1 - v/u)(1 + v/u)
Now, with v = V - u and sub it into (1):
mu = mv + MV
mu = mv + M(u + v)
mu = mv + Mu + Mv
(m + M)v = (m - M)u
v/u = (m - M)/(m + M)
So the fractional loss is:
(1 - v/u)(1 + v/u) = [1 - (m - M)/(m + M)][1 + (m - M)/(m + M)]
= [2M/(m + M)][2m/(m + M)]
= 4mM/(m + M)2
2009-11-15 16:14:44 補充：
[(m-M)/(m+M)]^2 - 1 gives - 4mM/(m + M)^2 which is negative and not reasonable.
參考: Myself
2009-11-15 11:40 pm
(a) By the Conservation of Momentum,
MV + mv = mu
By the Conservation of Energy,
(1/2) MV^2 + (1/2) mv^2 = (1/2) mu^2
MV^2 + mv^2 = mu^2
(b) Fractional Loss of kinetic energy of A.
((1/2) mv^2 - (1/2) mu^2) / (1/2)(mu^2)
= (mv^2)/(mu^2) -1
= v^2 / u^2 -1
= (v/u) ^2 -1
In (a),
MV + mv = mu
As v = V-u,
M(v+u) +mv = mu
(M+m)v + Mu= mu
(M+m)v = (m-M) u
v/u = (m-M)/(M+m)
Hence, fractional loss,
((m-M)/(m+M)) ^2 -1
MV + mv = mu
By the Conservation of Energy,
(1/2) MV^2 + (1/2) mv^2 = (1/2) mu^2
MV^2 + mv^2 = mu^2
(b) Fractional Loss of kinetic energy of A.
((1/2) mv^2 - (1/2) mu^2) / (1/2)(mu^2)
= (mv^2)/(mu^2) -1
= v^2 / u^2 -1
= (v/u) ^2 -1
In (a),
MV + mv = mu
As v = V-u,
M(v+u) +mv = mu
(M+m)v + Mu= mu
(M+m)v = (m-M) u
v/u = (m-M)/(M+m)
Hence, fractional loss,
((m-M)/(m+M)) ^2 -1
參考: /
2009-11-15 10:49 pm
(a) Conservation of momentum:
mu = mv + MV
Conservation of energy
(1/2)mu^2 = (1/2)mv^2 + (1/2)M.V^2
(b) From the momentum eqution,
M = (mu-mv)/M
substitute it into the energy equation and simplify,
v^2 - V(v) + (uV-u^2) = 0
The solution of this quadratic equation is
v = V-u or v = u (rejected because of impossible physical situation)
or v/u = V/u - 1
The fractional change of kinetic energy
=[ (1/2)mv^2 - (1/2)mu^2]/(1/2)mu^2
= (v^2-u^2)/u^2 = (v/u)^2-1
From the momentum equation,
m = m(v/u) + M(V/u)
m = m(v/u) + M(v/u + 1)
(v/u)(M+m) = m-M
v/u = (m-M)/(M+m)
Hence, fractional loss of kinetic energy
= [(m-M)/(M+m)]^2-1
mu = mv + MV
Conservation of energy
(1/2)mu^2 = (1/2)mv^2 + (1/2)M.V^2
(b) From the momentum eqution,
M = (mu-mv)/M
substitute it into the energy equation and simplify,
v^2 - V(v) + (uV-u^2) = 0
The solution of this quadratic equation is
v = V-u or v = u (rejected because of impossible physical situation)
or v/u = V/u - 1
The fractional change of kinetic energy
=[ (1/2)mv^2 - (1/2)mu^2]/(1/2)mu^2
= (v^2-u^2)/u^2 = (v/u)^2-1
From the momentum equation,
m = m(v/u) + M(V/u)
m = m(v/u) + M(v/u + 1)
(v/u)(M+m) = m-M
v/u = (m-M)/(M+m)
Hence, fractional loss of kinetic energy
= [(m-M)/(M+m)]^2-1
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